Question

Under the action of a force, a body of mass $4\text{kg}$ moves along $x-$ axis, such that $x$ as a function of time is given by $x=\frac{t^3}{3}$, where all quantities are in $\text{SI}$ units. What is the work done in the first three seconds ?

• A. $324\text{J}$
• B. $162\text{J}$
• C. $81\text{J}$
• D. $9\text{J}$

Answer 1

The correct option is B $162\text{J}$

Given: $x=\frac{t^3}{3}$

So, acceleration of the body will be

$a=\frac{d^{2}x}{d{t}^2}=2t{\text{m/s}}^2$

From Newton's second law of motion,

$F=ma=4\times 2t=8t\text{N}$

Work done by force $F$ is given by

$dW=Fdx=\left(8t\right)dx......\left(1\right)$

As, $v=\frac{dx}{dt}=t^2$

$\therefore dx=t^{2}dt$

Putting value of $dx$ in eq. $\left(1\right)$, we get

$dW=8t\left(t^{2}dt\right)=8t^{3}dt$

Therefore, work done in first $\text{three seconds}$ is

$W=\int dW=\underset{0}{\overset{3}{\int }}8t^{3}dt=8{\left[\frac{t^4}{4}\right]}_0^3$

$\therefore W=2(3{)}^4=162\text{J}$

Hence, option $\left(B\right)$ is correct.